What is the speed of the block just before it hits the spring?

Question

A 0.271-kg block is dropped straight downwards onto a vertical spring. The spring constant of the spring is 53 N/m. The block sticks to the spring and the spring compresses 0.15 m before coming to a momentary halt. What is the speed of the block just before it hits the spring?

Answer 1

Don't things fall at 32ft/sec.

Answer 2

The potential energy gained in the spring is

1/2 *spring constant * Compression^2

= 1/2 * 53 * 0.15

The loss of potential energy of the block between hitting the spring and coming to a momentary halt
= m g h = 0.271 * 9.8 * 0.15

The kinetic energy of the block when it hits the spring
1/2 m v^2 = (1/2) * 0.271 * v^2

Loss of energy of the block = gain in Potential energy of the spring

>>
(1/2) * 0.271 * v^2 + 0.271 * 9.8 * 0.15 = 1/2 * 53 * 0.15
>>
0.1355V^2 + 0.39837 = 7.1815
>>
v ~ 7.075 m/sec

Answer 3

you can not deduce the speed of the block until you have the distance the block will fall, from rest to the spring.

I'm assuming that the medium the block will be falling is not water or liquid.

You might also want to factor in air resistance!

Basically, multiply the distance by the time!

It doesn't matter what the spring constant is or that it will come to a halt, if you want the speed before it contacts the spring!

Answer 4

find the work consumed by the spring. subtract the amount of potential energy consumed in the compression of the spring. The remainder is the kinetic energy of the block

Answer 5

F = kx - mg = ma
a = (k/m)x - g = dv/dxdx/dt = v dv/dx
∫vdv = (k/m)∫xdx - ∫gdx

0 - (1/2)v² = (k/m)-(1/2)x² + gx)
v^2 = (-2*53/0.271)(-0.5(-0.15)^2 - (-0.15)(-9.80662)
v^2 = 4.400369 - 1.470993
v^2 = 2.29376 m²/s²
v = 1.711542 m/s